Electric field between two charges in 1D

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SUMMARY

The discussion focuses on calculating the electric field at point A, located 0.25 m to the right of a left-hand positive charge (+Q) in a 1D configuration with two identical positive charges separated by 1 m. The electric field (E) is determined using the formula E = kQ/r², where k is Coulomb's constant. The correct approach involves calculating the electric fields due to both charges separately and then vectorially adding them, considering their directions. The electric field at point A points to the right, as the force on a positive test charge is influenced by both charges.

PREREQUISITES
  • Understanding of Coulomb's Law and electric fields
  • Familiarity with vector addition of forces
  • Knowledge of the concept of electric field direction
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the concept of electric field lines and their representation
  • Learn about superposition principle in electric fields
  • Explore the implications of charge configurations on electric fields
  • Investigate the effects of distance on electric field strength
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Students studying electromagnetism, physics educators, and anyone interested in understanding electric fields and forces between charges.

Tekee
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Homework Statement



Two identical positive charges of +Q are 1 m apart. What is the magnitude and direction of the electric field at point A, 0.25 m to the right of the left-hand charge?

Homework Equations



E = kQ/r^2

The Attempt at a Solution



I summed up kQ+/(.25^2) and kQ+/(.75^2) and what I got was one of the answer choices, but apparently it was wrong. Evidently you have to subtract the .75 charge from the .25 to get the right answer. I also am unsure of which direction the electric field points.
 
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okay tekee, imagine that the point of interest is a positive charge. Which should be the default. A positive charge to the righthand side of it pushes it left. and a positive charge to the lefthand side of it will push right. So you will have to make appropriate changes to your answer due to that.
 
point A is between the charges. The direction of the electric field is the direction of the force on a positive test-charge.
What is the magnitude and direction of the field due to the positive charge 0.25 m to the left of A?
What is the magnitude and direction of the field due to the positive charge 0.75 m to the right of A?
Now you can add those fields like vectors.
 

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